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Scalars and Vectors

Scalars and Vectors. A scalar quantity is one that can be described by a single number: temperature, speed, mass. A vector quantity deals inherently with both magnitude and direction : velocity, force, displacement, acceleration. Scalars and Vectors.

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Scalars and Vectors

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  1. Scalars and Vectors A scalar quantity is one that can be described by a single number: temperature, speed, mass A vector quantity deals inherently with both magnitude and direction: velocity, force, displacement, acceleration

  2. Scalars and Vectors Arrows are used to represent vectors. The direction of the arrow gives the direction of the vector. By convention, the length of a vector arrow is proportional to the magnitude of the vector. 8 lb 4 lb

  3. Vectors The plane travels with a velocity relative to the ground which is vector sum of the plane’s velocity (relative to the air) plus the wind velocity The resultant is a combination of both motions

  4. Vector Procedures These are the rules you must play by to add vectors and find the resultant: Draw each vector to scale and label Vectors should be drawn in order as they occurred and labeled (I.e.v1, v2…..) Draw each successive vector starting from the tip of the preceding vector. (This is called the “tip to tail” method The vector sum is found by drawing the resultant vector arrow from the tail of the first vector to the tip of the last vector For two-vector operations in two dimensions the resultant forms the diagonal of a parallelogram with the component vectors forming the sides. v1 Vector v1 is 1 unit east or +1 unit A1 v2 A2 Vector v2 is 3 units west or - 3 units Vector (v1 + v2) is 2 units west or - 2 units R = A1 + A2 v1 + v2 = R  v1 v2 R = Runits @ 0 ( measured from East counterclockwise)

  5. Use a ruler and vector rules to find the resultant of the following vector additions. Record your answer using vector convention. R = F1+ F2 F1 F2 F1 F2 P1 P1 P2 R = P1+ P2 P2 R = P2+ P1 P1

  6. Vector Addition and Subtraction When a vector is multiplied by -1, the magnitude of the vector remains the same, but the direction of the vector is reversed.

  7. Vector Addition and Subtraction

  8. Finding the components of a vector To find the magnitude of a vector quantity in a particular direction you can use the parallelogram rule Draw the vector and treat it as the resultant Form a parallelogram around the resultant in the two directions of interest. In physics problems these are usually perpendicular directions to simplify the problem Draw the component vectors along the sides of the parallelogram ah Pythagorean Theorem: a2 = ah2 + av2 av a a Eparalel E E Pythagorean Theorem: E2 = Eparalel2 + Eperpendicular2 Eperpendicular

  9. Using Trig to solve 2D-Vector Problems Right-Angled Trig Relationships Trig functions for right-angled triangles are just geometric ratios. You probably memorized these ratios in your math classes using So/h..Ca/h..To/a. Here they are once more! sin = o / h cos = a / h h o tan = sin / cos = o / a  a Follow these rules! Sketch vectors on a coordinate system. (Two perpendicular axes) Find (resolve) the components of all vectors in the two directions of interest using the parallelogram rule. Make sure that you use the normal conventions of + and – arithmetic quantities. Sum the components in the two directions of interest. Find the magnitude of R using Pythagorean theorem Use the tangent rule to determine the direction R as measured from the East by convention or a specified direction

  10. Practice, Practice…..Practice Use trig relationships to determine i) the other component and ii) the resultant R i) tan = vv / vh R vv vv = vhtan  vv = (35 m/s) tan 600 600 vh= 35 m/s vv = 60.6 m/s ii) R =  (vv2 + vh2) R =  ((60.6 m/s)2 + (35 m/s)2) R = 70 m/s  = 1800 - 600 = 1200 R = 70 m/s @ 1200

  11. Determine the resultant R of the following vectors by using the horizontal and vertical components of A and B R = 100 Newtons A cos 600 A sin 600 A = 80 Newtons + -80 N cos 600 +80 N sin 600 B= 60 Newtons - 40 N + 69.3 N Av Bv B cos 300 B sin 300  600 300 - + 60 N sin 300 +60 N cos 300 AH BH R =  (RH2 + Rv2) + 52 N + 30 N R =  ((12 N)2 + (99.3 N)2) + 12 N + 99.3 N R = 100 N  = tan-1 (Rv / RH)  = tan-1 (99.3 N / 12 N) R = 100 N @ 830  = 830

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